If abc = 1, show that
Question:

If abc = 1, show that

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

Solution:

To prove,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

$=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$

$=\frac{b}{b+a b+1}+\frac{c}{c+b c+1}+\frac{a}{a+a c+1} \ldots .$(1)

We know abc = 1

c = 1/ab

By substituting the value c in equation (1), we get

$=\frac{\mathrm{b}}{\mathrm{b}+\mathrm{ab}+1}+\frac{\frac{1}{\mathrm{ab}}}{\frac{1}{\mathrm{ab}}+\mathrm{b}\left(\frac{1}{\mathrm{ab}}\right)+1}+\frac{\mathrm{a}}{\mathrm{a}+\mathrm{a}\left(\frac{1}{\mathrm{ab}}\right)+1}$

$=\frac{b}{b+a b+1}+\frac{1}{1+b+a b}+\frac{a b}{1+a b+b}$

$=\frac{1+a b+b}{b+a b+1}$

$=1$

Therefore, LHS = RHS Hence proved